Indefinite Integrals

$\begin{aligned} I & = \int_1^4 \frac {2+\sqrt x}{(x+1+\sqrt x)^2} dx & \small \color{#3D99F6} \text{Let }u = \sqrt x \implies du = \frac {dx}{2\sqrt x} \\ & = \int_1^2 \frac {2u(u+2)}{(u^2+u+1)^2} du & \small \color{#3D99F6} \text{See note.} \\ & = - \frac {2(u+1)}{u^2+u+1} \bigg|_1^2 \\ & = - \frac 67 + \frac 43 \\ & = \frac {10}{21} \end{aligned}$

Therefore, $\beta - \alpha = 21-10 = \boxed{11}$ .

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Note: Assuming the integral is of the form::

$\begin{aligned} I & = \frac {au+b}{u^2+u+1} + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \frac {dI}{du} & = \frac {a(u^2+u+1)-(au+b)(2u+1)}{(u^2+u+1)^2} \\ \frac {2u(u+2)}{(u^2+u+1)^2} & = \frac {-au^2-2bu+a-b}{(u^2+u+1)^2} & \small \color{#3D99F6} \text{Equating the coefficients} \\ \implies a & = b = -2 \\ \implies I & = - \frac {2(u+1)}{u^2+u+1} \end{aligned}$

Using quotient rule we see the desired indefinite is $\frac{2x}{x+\sqrt{x}+1}+C$ . So answer is 11